Question

g What is the final velocity of a hoop that rolls without slipping down a 6.92 m high hill, starting from rest

Answer 1

Answer:

The value is  $v = 8.24 \ m/s$

Explanation:

From the question we are told that

  The height of the hill is  $h = 6.92 \ m$

Generally from the law of energy conservation we have that

   $PE = KE + RKE$

Here  PE is the potential energy of the hoop which is mathematically represented as

       $PE = mgh$

 KE is the kinetic energy of the hoop  which is mathematically represented as

     $KE = \frac{1}{2} * m * v^2$

And  

  RKE  is the rotational  kinetic energy which is mathematically represented as

      $RKE = \frac{1}{2} * I * w^2$

Here I  is the moment of inertia of the hoop which is mathematically represented as

           $I = m * r^2$

and  w  is the angular velocity which is mathematically represented as

          $w = \frac{v}{r}$

So

         $RKE = \frac{1}{2} * m r^2 * \frac{v}{r} ^2$

=>     $RKE = \frac{1}{2} * m r * v^2$

So

     $mgh = \frac{1}{2} * m * v^2 + \frac{1}{2} * m r * v^2$

=>   $v = \sqrt{gh}$

=>   $v = \sqrt{9.8 * 6.92 }$

=>   $v = 8.24 \ m/s$